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Topic: Sum of primes divided by the next prime
Replies: 6   Last Post: Aug 2, 2006 4:58 PM

 Messages: [ Previous | Next ]
 Tapio Hurme Posts: 853 Registered: 12/8/04
Re: Sum of primes divided by the next prime
Posted: Aug 2, 2006 4:23 PM

You are right! AFAICS

Tapio

"Tim Peters" <tim.one@comcast.net> wrote in message
news:beednTIHXN9BnkzZnZ2dnUVZ_vSdnZ2d@comcast.com...
> [Tapio]
>> Sum the primes up to p_n. Divide the sum by the next prime p_(n+1):
>>
>> Sum (2+3+...+p_n)/(p_(n+1))
>>
>> I found with the aid of excel two examples:
>>
>> (2+3) = 0 mod 5, i.e. the next prime 5 divides the sum of all the
>> preceding primes.

>
> While degenerate, it would also be reasonable to say that 2 divides 0.
>

>> (2+3+....+61) = 0 mod 67
>
> Not when I do it ;-)
>
> sum(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61) =
> 501 =
> 7*67 + 32
>
> However, I do get that 71 divides 568.
>

>> Below the prime 10009 there is no other cases, except those mentioned
>> above.
>> Are there any other p_n > 10009 ?

>
> Through the 50,847,534 primes less than a billion (10^9), the only cases I
> found were:
>
> 2 divides 0
> 5 divides 5
> 71 divides 568
> 369119 divides 5536785000
> 415074643 divides 4456254649636576
>
> Best guess is that mod(sum(p_i, i = 1 to n), p_(n+1)) is "essentially
> random", so that the chance of seeing 0 is 1/p_(n+1). Then since
> sum(1/p_i) is divergent (albeit very slowly), best guess is that the
> number of primes for which the mod is 0 is infinite (although their
> density will decrease rapidly). Suggestive:
>
> p_n sum(1/p for prime p in [2, p_n])
> ----- --------------------------------
> 5 1.03
> 71 1.74
> 369119 2.81
> 415074643 3.25
>
>

Date Subject Author
8/2/06 Tapio Hurme
8/2/06 Jan Kristian Haugland
8/2/06 Tapio Hurme
8/2/06 mensanator
8/2/06 Tapio Hurme
8/2/06 Tim Peters
8/2/06 Tapio Hurme